的Oracle SQL相关子查询 - 在一些列
问题描述:
返回COUNT(*)我有我最初的声明是:的Oracle SQL相关子查询 - 在一些列
SELECT TEAM.ID PKEY_SRC_OBJECT,
TEAM.MODF_DAT UPDATE_DATE,
TEAM.MODF_USR UPDATED_BY,
PERSO.FIRST_NAM FISRT_NAME
FROM TEAM
LEFT OUTER JOIN PERSO ON (TEAM.ID=PERSO.TEAM_ID)
我想计算一些“标志”和我最初的发言归还。 有3个标志,其可以这样来计算:
1)标志ISMASTER:
SELECT Count(*)
FROM TEAM_TEAM_REL A, TEAM B
WHERE B.PARTY_PTY_ID = A.RLTD_TEAM_ID
AND CODE = 'Double';
2)标志ISAGENT:
SELECT Count(*)
FROM TEAM_ROL_REL A, TEAM B
WHERE B.PARTY_PTY_ID = A.TEAM_ID;
3)标志NUMPACTS:
SELECT Count(*)
FROM TEAM_ROL_REL A,
TEAM_ROL_POL_REL B,
PERSO_POL_STA_REL C,
TEAM D
WHERE A.ROL_CD IN ('1','2')
AND A.T_ROL_REL_ID = B.P_ROL_REL_ID
AND B.P_POL_ID = C.P_POL_ID
AND C.STA_CD = 'A'
AND D.PARTY_PTY_ID = A.TEAM_ID;
为了达到这个目的,我已经更新了我的初始语句:
WITH ABC AS (
SELECT TEAM.ID PKEY_SRC_OBJECT,
TEAM.MODF_DAT UPDATE_DATE,
TEAM.MODF_USR UPDATED_BY,
PERSO.FIRST_NAM FISRT_NAME
FROM TEAM
LEFT OUTER JOIN PERSO ON (TEAM.ID=PERSO.TEAM_ID)
)
SELECT ABC.*, MAST.ISMASTER, AGENT.ISAGENT, PACTS.NUMPACTS FROM ABC
LEFT OUTER JOIN (
select
RLTD_TEAM_ID,
Count(RLTD_TEAM_ID) OVER (PARTITION BY RLTD_TEAM_ID) as ISMASTER
FROM TEAM_TEAM_REL
WHERE CODE = 'Double'
) MAST
ON ABC.PKEY_SRC_OBJECT = MAST.RLTD_TEAM_ID
LEFT OUTER JOIN (
select
TEAM_ID,
Count(TEAM_ID) OVER (PARTITION BY TEAM_ID) as ISAGENT
FROM TEAM_ROL_REL
) AGENT
ON ABC.PKEY_SRC_OBJECT = AGENT.TEAM_ID
LEFT OUTER JOIN (
select
TEAM_ID,
Count(TEAM_ID) OVER (PARTITION BY TEAM_ID) as NUMPACTS
FROM TEAM_ROL_REL, TEAM_ROL_POL_REL, PERSO_POL_STA_REL
WHERE TEAM_ROL_REL.ROL_CD IN ('1','2')
AND TEAM_ROL_REL.T_ROL_REL_ID = TEAM_ROL_POL_REL.P_ROL_REL_ID
AND TEAM_ROL_POL_REL.P_POL_ID = PERSO_POL_STA_REL.P_POL_ID
AND PERSO_POL_STA_REL.STA_CD = 'A'
) PACTS
ON ABC.PKEY_SRC_OBJECT = PACTS.TEAM_ID;
对于两个第一标志(ISMASTER和ISAGENT)我得到的结果,在不到1分钟,但最后的标志(NUMPACTS)运行几分钟,而不提供任何结果。 我认为我的发言太重了,也许我应该以完全不同的方式去做。
答
我想你或许有过复杂的事情。
你可以这样做(假设我理解正确的您的要求),像这样:
WITH ttr AS (SELECT rltd_team_id,
COUNT(*) is_master
FROM team_team_rel
AND CODE = 'Double'
GROUP BY rltd_team_id),
trr AS (SELECT team_id,
COUNT(*) is_agent
FROM team_rol_rel
GROUP BY team_id)
pacts AS (SELECT trr1.team_id,
COUNT(*) num_pacts
FROM team_rol_rel trr1
INNER JOIN team_rol_pol_rel trpr ON (trr1.t_rol_rel_id = trpr.p_rol_rel_id)
INNER JOIN perso_pol_sta_rel ppsr ON (trpr.p_pol_id = ppsr.p_pol_id
WHERE trr1.rol_cd IN ('1', '2')
AND ppsr.st_cd = 'A'
GROUP BY trr1.team_id)
SELECT t.id pkey_src_object,
t.modf_dat update_date,
t.modf_usr updated_by,
p.first_nam first_name,
ttr.is_master,
trr.is_agent,
pacts.num_pacts
FROM team t
LEFT OUTER JOIN perso p ON t.id = p.team_id
LEFT OUTER JOIN ttr ON t.party_pty_id = ttr.rltd_team_id
LEFT OUTER JOIN trr ON t.party_pty_id = trr.team_id
LEFT OUTER JOIN pacts ON t.pkey_src_object = pacts.team_id;
注:未经测试,因为您没有提供任何测试数据。
+1
这样工作很好,谢谢! – baboufight
运行解释计划并发布结果。 – OldProgrammer